Verified Solution • 2025–26 Session moderate

QJEEAdv15Chaf69: JEE Advanced 2015 Chemistry - Class 11 JEEAdv15Chaf69 Solution: One mole of a monoatomic real gas satisfies the equation p(V b) …

One mole of a monoatomic real gas satisfies the equation p(V −- b) = RT where b is a constant. The relationship of interatomic potential V(r) and interatomic distance r for the gas is given by

Multiple Choice Options

  • A https://ik.imagekit.io/novatra/photos/296d5c4bce0eb58b97c7c78ba4b1650cb5ea066ccd8a429ce6c647105c82ca6d.webp
  • B https://ik.imagekit.io/novatra/photos/c916ab0d4da7cd14abb72752131716937c684ed1d7d77341758eae0eae70c8f7.webp
  • C https://ik.imagekit.io/novatra/photos/6b43b713572c2fc80826a61145a37976bcc15e82cdf3117fec3e55640c55a0a1.webp
  • D https://ik.imagekit.io/novatra/photos/26037a112cc41d15ce9dc8ebca6a67236b6cc6b5b596640a92cf932840f5e8d0.webp

Step-by-Step Verified Solution

The given equation is P(V−b)=RT\mathrm{P}(\mathrm{V}-b)=\mathrm{RT} ,

On comparing with van Der Waal's equation

[P+a V2][V−b]=RT we get a=0\left[\mathrm{P}+\frac{a}{\mathrm{~V}^2}\right][\mathrm{V}-b]=\mathrm{RT} \text { we get } a=0

Hence, only repulsive forces are present which are contributive only at very close distance.

Thus, the potential energy will increase abruptly, so graph (c) is correct. The dominance of repulsive force can be shown by using the compressibility factor.

P(V−b)amp;=RTPVamp;=Pb+RTPVRTamp;=PbRT+1\begin{aligned}\mathrm{P}(\mathrm{V}-b) & =\mathrm{RT} \\\\\mathrm{PV} & =\mathrm{Pb}+\mathrm{RT} \\\\\frac{\mathrm{PV}}{\mathrm{RT}} & =\frac{\mathrm{Pb}}{\mathrm{RT}}+1\end{aligned}

Z=\frac{P b}{\mathrm{RT}}+1 \text {, i.e., } \mathrm{Z}>1 \text { (Repulsive forces) }

Related Learning Resources & Exam Preparation

JEE Advanced PYQ Question Bank
Browse year-wise authentic previous years' questions with solutions
Authentic CBT Mock Tests
Simulate official exam environment with live timer and percentile scoring